Probability and entanglement evolutions for Szegedy’s quantum search on the one-dimensional cycle with self-loops


The Szegedy’s quantum walk can give rise to a quadratic speed-up when the Markov chain is ergodic and symmetric. However, the quantum search on a one-dimensional (1D) cycle graph does not achieve a speed-up. In this paper, we study the effects of self-loops on the 1D cycle by Szegedy’s quantum search. First, with the help of self-loops, Szegedy’s quantum search can increase the success probability of finding a marked vertex on the 1D cycle. Second, the general expressions for the evolving states and the success probability on the 1D cycle with self-loops are explicitly presented by the symmetric tridiagonal matrix. The evolution of success probability is slower and smaller with the increase in the weight of the self-loops. Third, an approximate entanglement formula of the success probability is derived by a concave function, where the entanglement is measured by the reduced von Neumann entropy. The existence of a turning point is confirmed, and it was found to depend on the maximum eigenvalue of the initial superposition state and the number of marked vertices. Before the turning point, the entanglement first increased and then decreased.

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This work is supported by the National Natural Science Foundation of China (Grant Nos. 61871120 and 61502101), Natural Science Foundation of Jiangsu Province, China (Grant No. BK20191259), the Six Talent Peaks Project of Jiangsu Province (Grant No. XYDXX-003), and the Fundamental Research Funds for the Central Universities.

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Correspondence to Zhihao Liu or Hanwu Chen.

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Appendix A Proof of Eq. (59)

When \(m = 1,k = 1\) in Eq. (27), according to the definition of the integral, we can derive

$$\begin{aligned}&\sum \limits _{j = 1}^{n - 1} {\sin \frac{\pi }{n} +\sin \frac{j\pi }{n}+ \cdots + \sin \frac{{\left( {n - 1} \right) \pi }}{n}} \\&\quad =\frac{n}{\pi }\sum \limits _{i = 2}^n {\frac{\pi }{n}\sin \frac{{\left( {i - 1} \right) \pi }}{n}} = \frac{n}{\pi }\int _0^\pi {\sin x} \text {d}x = \frac{{2n}}{\pi },\\&f = {\sum \limits _{j = 1}^{n - 1} {{{\left( {\sin \frac{\pi }{n}} \right) }^2} +{{\left( {\sin \frac{j\pi }{n}} \right) }^2}+ \cdots + \left( {\sin \frac{{\left( {n - 1} \right) \pi }}{n}} \right) } ^2} \\&\quad = \frac{n}{\pi }\int _0^\pi {{{\left( {\sin x} \right) }^2}} \text {d}x = \frac{n}{2}. \end{aligned}$$

When \(m = 1,k > 1\) and k is odd,

$$\begin{aligned}&\sum \limits _{j = 1}^{n - 1} {\sin \frac{{k\pi }}{n} +\sin \frac{jk\pi }{n} \cdots + \sin \frac{{2\left( {n - 1} \right) k\pi }}{n}} \\&\quad \le \frac{n}{{k\pi }}\sum \limits _{i = 2}^n {\frac{{k\pi }}{n}\sin \frac{{\left( {i - 1} \right) k\pi }}{n}} = \frac{{2n}}{{k\pi }}, \end{aligned}$$

where f does not change. So

$$\begin{aligned} \frac{{\sum \limits _k {\left( {\sum \limits _{j = 1}^{n - 1} {\sin \left( {\frac{{jk\pi }}{n}} \right) } } \right) } }}{{\sqrt{n} {{\left( {\sqrt{f} \sqrt{2} \sin {\theta _k}} \right) }^2}}} \le \sum \limits _k {\frac{2}{{k\pi {{\left( {\sin {\theta _k}} \right) }^2}\sqrt{n} }}}. \end{aligned}$$

Case 1: for the coefficient of \(\left| {{c_{1,1}}} \right| \),

$$\begin{aligned}&\left| {{c_{1,1}}} \right| \le \left| \sum \limits _k {\frac{2}{{k\pi {{\left( {\sin {\theta _k}} \right) }^2}\sqrt{n} }}\sqrt{b} } \sin \frac{{k\pi }}{n}2\cos \left( {2{\theta _k}t} \right) 2\left( {1 - \cos {\theta _k}} \right) \right| \\&\quad \le \left| \frac{{2\sqrt{b} }}{{\sqrt{n} }}\sum \limits _k {\frac{{4\cos \left( {2{\theta _k}t} \right) }}{{k\pi \left( {1 + \cos {\theta _k}} \right) }}} \frac{{k\pi }}{n}\right| . \end{aligned}$$

Since \( - 2 \le \frac{{4\cos \left( {2{\theta _k}t} \right) }}{{\left( {1 + \cos {\theta _k}} \right) }} \le 2\), we have \(\left| {{c_{1,1}}} \right| < \frac{{2\sqrt{b} }}{{n\sqrt{n} }} \times 2 \times \frac{n}{2} = \frac{{2\sqrt{b} }}{{\sqrt{n} }}\). So

$$\begin{aligned} \left| {{c_{1,1}}} \right| \le \frac{{\sqrt{2b} }}{{\sqrt{n} }}. \end{aligned}$$

Case 2: for \(\left| {{c_{n,1}}} \right| \),

$$\begin{aligned}&\left| {{c_{n,1}}} \right| \le \left| {\sum \limits _k {\frac{2}{{k\pi {{\left( {\sin {\theta _k}} \right) }^2}\sqrt{n} }}\sqrt{a} } \sin \frac{{k\pi }}{n}\left( {\cos \left( {2{\theta _k}t} \right) - \cos \left( {2{\theta _k}t + {\theta _k}} \right) } \right) } \right| \\&\quad < \left| \frac{{\sqrt{a} }}{{\sqrt{n} }}\sum \limits _k {\frac{{{\theta _k}}}{{k\pi \left( {\left( {1 - \cos 2{\theta _k}} \right) /2} \right) }}} \frac{{k\pi }}{n}\right| , \end{aligned}$$

where \(\left| {\cos \alpha - \cos \beta } \right| \le \left| {\alpha - \beta } \right| \).

Since \(1 - \cos \left( {{\theta _k}} \right) \ge \frac{{{\theta _k}^2}}{8}\), then \(1 - \cos \left( {2{\theta _k}} \right) \ge \frac{{{\theta _k}^2}}{2}, \left( {0 \le {\theta _k} \le \frac{\pi }{2}} \right) \).

Then, \(\left| {{c_{n,1}}} \right| \le \frac{{\sqrt{a} }}{{\sqrt{n} }}\sum \limits _k {\frac{2}{{\left( {{\theta _k}} \right) }}} \frac{1}{n}\). Due to \({\theta _k} \approx \frac{k}{n}\) and k is odd, we can obtain approximately \(\left| {{c_{n,1}}} \right| \le \frac{{\sqrt{a} }}{{\sqrt{n} }}\sum \limits _k {\frac{1}{k}}\) (k is odd), where \(\sum \limits _k {\frac{1}{k}}\) is the harmonic series. When n is finite, the sum of the harmonic series can be calculated. Thus,

$$\begin{aligned} \left| {{c_{n,1}}} \right| \le \frac{{2\sqrt{a} }}{{\sqrt{n} }}. \end{aligned}$$

Case 3: for \(\left| {{c_{n,n}}} \right| \),

$$\begin{aligned} \left| {{c_{n,n}}} \right| = \frac{{\sqrt{b} }}{{\sqrt{n} }}. \end{aligned}$$

Case 4: for \(\left| {{c_{2,1}}} \right| \),

$$\begin{aligned}&\left| {{c_{2,1}}} \right| \le \left| \sum \limits _k {\frac{{2 \times 2\sqrt{a} }}{{k\pi {{\left( {\sin {\theta _k}} \right) }^2}\sqrt{n} }}} \left( \begin{array}{l} \sin \frac{{2k\pi }}{n}\left( {\left( {1 - \cos {\theta _k}} \right) \cos \left( {2{\theta _k}t} \right) - \sin {\theta _k}\sin \left( {2{\theta _k}t} \right) } \right) \\ + \sin \frac{{k\pi }}{n}\left( {\left( {1 - \cos {\theta _k}} \right) \cos \left( {2{\theta _k}t} \right) + \sin {\theta _k}\sin \left( {2{\theta _k}t} \right) } \right) \end{array} \right) \right| \\&\quad \approx \left| \sum \limits _k {\frac{2}{{k\pi {{\left( {\sin {\theta _k}} \right) }^2}\sqrt{n} }}2\sqrt{a} } \left( {2\sin \frac{{k\pi }}{n}\left( {\left( {1 - \cos {\theta _k}} \right) \cos \left( {2{\theta _k}t} \right) } \right) } \right) \right| . \end{aligned}$$

It is the same as \(\left| {{c_{1,1}}} \right| \), then

$$\begin{aligned} \left| {{c_{2,1}}} \right| \le \frac{{\sqrt{2a} }}{{\sqrt{n} }}. \end{aligned}$$

Case 5: for \(\left| {{c_{1,n}}} \right| \),

$$\begin{aligned} \left| {{c_{1,n}}} \right| = \left| {{c_{2,1}}} \right| \le \frac{{\sqrt{2a} }}{{\sqrt{n} }}. \end{aligned}$$

Thus, \({r_1} = \left| {{c_{n,1}} \cdot {c_{1,1}}} \right| + \left| {{c_{n,n}} \cdot {c_{1,n}}} \right| + \left| {{c_{n,1}} \cdot {c_{2,1}}} \right| < \frac{{2\sqrt{2ab} + \sqrt{2ab} + 2\sqrt{2} a}}{n}\), so

$$\begin{aligned} \sqrt{2} {r_1} < \frac{{6\sqrt{ab} + 4a}}{n}. \end{aligned}$$

Compared to \({\lambda _{0\max }} = \frac{{ 4\sqrt{ab} + 6a}}{n}\) in Eq. (39), we can conclude \(\sqrt{2}{r_1} \le {\lambda _{0\max }}\) in Eq. (59).

When \(m \ge 2\), these results still hold by substituting \(m\ge 2\) into \(m=1\).

Appendix B Proof of Eq. (60)

Similar, \(m \ge 1\), when \(1 \le j \le n - m\), \(\left| {{c_{j,j}}} \right| \approx \left| {{c_{\mathrm{{1,1}}}}} \right| \le \frac{{\sqrt{2b} }}{{\sqrt{n} }}\), \(\left| {{c_{j,j + 1}}} \right| \approx \left| {{c_{\mathrm{{2,1}}}}} \right| \le \frac{{\sqrt{2a} }}{{\sqrt{n} }}\). Thus,

$$\begin{aligned}&\left\| {{\rho ^X}} \right\| _1 = \frac{{4\left| {{c_{j,j}}{c_{j,j + 1}}} \right| + {c_{j,j}}^2 + 3{c_{j,j + 1}}^2 + \left| {{c_{n,1}}{c_{j,j + 1}}} \right| }}{n}\nonumber \\&\quad \le \frac{{2b + 6a + 8\sqrt{ab} + 2\sqrt{2} a}}{n}\nonumber \\&\quad \approx \frac{{2b + 6a + 8\sqrt{ab} }}{n}\nonumber \\&\quad = 2{\lambda _{0\max }}. \end{aligned}$$

Appendix C Proof of Eq. (69)

Similar, \(m \ge 2\), where

$$\begin{aligned} {r_1}^\prime&= \left| {{c_{n,1}}\cdot {c_{1,1}}} \right| + \left| {{c_{n,n}}\cdot {c_{1,n}}} \right| + \left| {{c_{n,1}}\cdot {c_{2,1}}} \right| + \left| {{c_{n,n - 1}}\cdot {c_{n - 2,n - 1}}} \right| \nonumber \\&\quad + \left| {{c_{n,n - 1}}\cdot {c_{n - 1,n - 1}}} \right| \nonumber \\&\quad + \left| {{c_{n,n}}\cdot {c_{n - 1,n}}} \right| \nonumber \\&\le {\lambda _{0\max }} + \frac{{\sqrt{2a} + 2\sqrt{ab} }}{n}\nonumber \\&<2{\lambda _{0\max }}. \end{aligned}$$

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Xu, M., Liu, Z., Chen, H. et al. Probability and entanglement evolutions for Szegedy’s quantum search on the one-dimensional cycle with self-loops. Quantum Inf Process 20, 51 (2021).

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  • Szegedy’s quantum search
  • Self-loops
  • Tridiagonal matrix
  • Entanglement measure
  • Turning point